Buffers Lecture Notes

The Danger of Antifreeze

• Each year, thousands of pets and wildlife species die from consuming antifreeze.
• Most brands of antifreeze contain ethylene glycol.
  • Sweet taste.
  • Initial effect is drunkenness.
• Ethylene glycol is metabolized in the liver to glycolic acid with the formula HOCH_2COOH.

Why Is Glycolic Acid Toxic?

• Glycolic acid, in high enough concentrations in the bloodstream, overwhelms the buffering ability of HCO_3^- in the blood, causing the blood pH to drop.
• Low blood pH compromises blood's ability to carry O_2 (Acidosis).
• The equilibrium is represented as: HbH^+(aq) + O2(g) \rightleftharpoons HbO2(aq) + H^+(aq).
• One treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol.

Buffer Solutions

• Resist changes in pH when an acid or base is added.
• Act by neutralizing acid or base that is added to the buffered solution.
• Contain either:
  • Significant amounts of a weak acid and its conjugate base.
  • Significant amounts of a weak base and its conjugate acid.
  • Blood has a mixture of H2CO3 and HCO_3^-.

Making an Acidic Buffer Solution

• It must contain significant amounts of both a weak acid and its conjugate base.

An Acidic Buffer Solution

• If a strong base is added, it is neutralized by the weak acid (HC2H3O_2) in the buffer.
  • NaOH(aq) + HC2H3O2(aq) \rightarrow H2O(l) + NaC2H3O_2(aq)
• If the amount of NaOH added is less than the amount of acetic acid present, the pH change is small.
• If a strong acid is added, it is neutralized by the conjugate base (NaC2H3O_2) in the buffer.
  • HCl(aq) + NaC2H3O2 \rightarrow HC2H3O2(aq) + NaCl(aq)
• If the amount of HCl is less than the amount of NaC2H3O_2 present, the pH change is small.

How Acid Buffers Work: Addition of Base

• HA(aq) + H2O(l) \rightleftharpoons A^-(aq) + H3O^+(aq)
• Buffers work by applying Le Châtelier’s principle to weak acid equilibrium.
• Buffer solutions contain significant amounts of the weak acid molecules, HA.
• These molecules react with added base to neutralize it: HA(aq) + OH^-(aq) \rightarrow A^-(aq) + H_2O(l).
• Alternatively, think of the H3O^+ combining with the OH^- to make H2O; the H_3O^+ is then replaced by the shifting equilibrium.

How Acid Buffers Work: Addition of Acid

• HA(aq) + H2O(l) \rightleftharpoons A^-(aq) + H3O^+(aq)
• The buffer solution also contains significant amounts of the conjugate base anion, A^-.
• These ions combine with added acid to make more HA: H^+(aq) + A^-(aq) \rightarrow HA(aq).
• After the equilibrium shifts, the concentration of H_3O^+ is kept constant.

Common Ion Effect

• HA(aq) + H2O(l) \rightleftharpoons A^-(aq) + H3O^+(aq)
• Adding a salt containing the anion NaA, which is the conjugate base of the acid (the common ion), shifts the position of equilibrium to the left.
• This causes the pH to be higher than the pH of the acid solution, lowering the H_3O^+ ion concentration.

Henderson–Hasselbalch Equation

• An equation derived from the K_a expression that allows us to calculate the pH of a buffer solution.
• The equation calculates the pH of a buffer from the pK_a and initial concentrations of the weak acid and salt of the conjugate base, as long as the “x is small” approximation is valid.

Deriving the Henderson–Hasselbalch Equation

• Consider the equilibrium expression for a generic acidic buffer.
• Solving for H_3O^+ gives:
  • Taking the logarithm of both sides and expanding log (AB) = log A + log B
  • Multiplying both sides by –1 and rearranging:
    • -log[H3O^+] = -logKa + log([A^-]/[HA])
  • Since pH = –log[H3O^+] and pKa = -logK_a
    • pH = pK_a + log([A^-]/[HA])

Equilibrium Analysis vs. Henderson–Hasselbalch Equation

• The Henderson–Hasselbalch equation is generally good enough when the “x is small” approximation is applicable.
• Generally, the “x is small” approximation will work when both of the following are true:
  • The initial concentrations of acid and salt are not very dilute.
  • The K_a is fairly small.
• For most problems, this means that the initial acid and salt concentrations should be over 100 to 1000 times larger than the value of K_a.

Calculating the pH of a Buffer Solution – Example

• Steps when calculating the pH of a buffer solution:
  1. Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing the given concentrations of the acid and its conjugate base as the initial concentrations. Leave room in the table for the changes in concentrations and for the equilibrium concentrations.
  2. Represent the change in the concentration of H_3O^+ with the variable x. Express the changes in the concentrations of the other reactants and products in terms of x.
  3. Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.
  4. Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization constant. In most cases, you can make the approximation that x is small.
  5. Substitute the value of the acid ionization constant into the K_a expression and solve for x. Confirm that x is small by calculating the ratio of x and the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). Therefore the approximation is valid.
  6. Determine the H_3O^+ concentration from the calculated value of x and substitute into the pH equation to find pH.

Examples and Practice Problems

• Example 16.1: Calculate the pH of a buffer solution that is 0.100 M in HC2H3O2 and 0.100 M in NaC2H3O2.
• For Practice 16.1: Calculate the pH of a buffer solution that is 0.200 M in HC2H3O2 and 0.100 M in NaC2H3O2.
• For More Practice 16.1: Calculate the pH of the buffer that results from mixing 60.0 mL of 0.250 M HCHO2 and 15.0 mL of 0.500 M NaCHO2.
• Example 16.2: Calculate the pH of a buffer solution that is 0.050 M in benzoic acid (HC7H5O2) and 0.150 M in sodium benzoate (NaC7H5O2). For benzoic acid, K_a = 6.5 × 10^{–5}.
  • Solve using both the equilibrium approach and the Henderson–Hasselbalch equation.
• For Practice 16.2: Calculate the pH of a buffer solution that is 0.250 M in HCN and 0.170 M in KCN. For HCN, Ka = 4.9 × 10^{–10} (pKa = 9.31).
  • Use both the equilibrium approach and the Henderson–Hasselbalch approach.

Change in pH After Adding Acid or Base

• Calculating the new pH after adding acid or base requires breaking the problem into two parts:
  • A stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other.
    • Added acid reacts with the A^- to make more HA.
    • Added base reacts with the HA to make more A^-.
  • An equilibrium calculation of [H_3O^+] using the new initial values of [HA] and [A^-].

Example: Calculating pH Change in a Buffer

• Example 16.3: A 1.0 L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8 × 10^{–5}. Because the initial amounts of acid and conjugate base are equal, the pH of the buffer is equal to pK_a = –log(1.8 × 10^{–5}) = 4.74. Calculate the new pH after adding 0.010 mol of solid NaOH to the buffer. For comparison, calculate the pH after adding 0.010 mol of solid NaOH to 1.0 L of pure water. (Ignore any small changes in volume that might occur upon addition of the base.)
  • Part I: Stoichiometry. The addition of the base converts a stoichiometric amount of acid to the conjugate base (adding base creates more base). Write an equation showing the neutralization reaction and then set up a table to track the changes.
  • Part II: Equilibrium. Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table. Use the amounts of acid and conjugate base from part I as the initial amounts of acid and conjugate base in the ICE table.
  • Substitute the expressions for the equilibrium concentrations of acid and conjugate base into the expression for the acid ionization constant. Make the x is small approximation and solve for x.
  • Calculate the pH from the value of x, which is equal to [H_3O^+].
  • Alternatively, use the Henderson–Hasselbalch equation, substituting the quantities of acid and conjugate base after the addition (from part I) into the Henderson–Hasselbalch equation and calculate the new pH.
• The pH of 1.0 L of water after adding 0.010 mol of NaOH is calculated from the [OH^–]. For a strong base, [OH^–] is simply the number of moles of OH^– divided by the number of liters of solution.
• Check: Adding base should make the solution more basic (higher pH); adding acid should make the solution more acidic (lower pH).
• For Practice 16.3: Calculate the pH of the solution in Example 16.3 upon addition of 0.015 mol of NaOH to the original buffer.
• For More Practice 16.3: Calculate the pH of the solution in Example 16.3 upon addition of 10.0 mL of 1.00 M HCl to the original buffer in Example 16.3.

Concept Check

• What change will be caused by the addition of a small amount of HCl to a HF/NaF buffer?
  • The correct answer is: [F–] will decrease and [HF] will increase.

Basic Buffers

• B:(aq) + H_2O(l) \rightleftharpoons H:B^+(aq) + OH^− (aq)
• Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B^+Cl^-.
  • H2O(l) + NH3 (aq) \rightleftharpoons NH_4^+(aq) + OH^− (aq)

Henderson–Hasselbalch Equation for Basic Buffers

• The Henderson–Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product.
• The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product.
  • B: + H_2O \rightleftharpoons H:B^+ + OH^−
• To apply the Henderson–Hasselbalch equation, the chemical equation of the basic buffer must be looked at like an acid reaction.
  • H:B^+ + H2O \rightleftharpoons B: + H3O^+
  • This does not affect the concentrations, just the way we are looking at the reaction.

Relationship between pKa and pKb

• Just as there is a relationship between the Ka of a weak acid and Kb of its conjugate base, there is also a relationship between the pKa of a weak acid and the pKb of its conjugate base.
  • Ka · Kb = K_w = 1.0 x 10^{−14}
  • -log(Ka · Kb) = −log(K_w) = 14
  • -log(Ka) + −log(Kb) = 14
  • pKa + pKb = 14

Henderson–Hasselbalch Equation in terms of pOH

• We can rewrite the Henderson–Hasselbalch equation for the chemical equation of the basic buffer in terms of pOH.

Example: Using Henderson–Hasselbalch Equation for Basic Buffers

• Example 16.4: Use the Henderson–Hasselbalch equation to calculate the pH of a buffer solution that is 0.50 M in NH3 and 0.20 M in NH4Cl. For ammonia, pK_b = 4.75.
  • Since Kb for NH3 (1.8 × 10^{–5}) is much smaller than the initial concentrations in this problem, you can use the Henderson–Hasselbalch equation to calculate the pH of the buffer.
  • First calculate pKa from pKb.
  • Then substitute the given quantities into the Henderson–Hasselbalch equation and calculate pH.
• For Practice 16.4: Calculate the pH of 1.0 L of the solution in Example 16.4 upon addition of 0.010 mol of solid NaOH to the original buffer solution.
• For More Practice 16.4: Calculate the pH of 1.0 L of the solution in Example 16.4 upon addition of 30.0 mL of 1.0 M HCl to the original buffer solution.

Buffering Effectiveness

• A good buffer should be able to neutralize moderate amounts of added acid or base.
• However, there is a limit to how much can be added before the pH changes significantly.
• The buffering capacity is the amount of acid or base a buffer can neutralize.
• The buffering range is the pH range the buffer can be effective.
• The effectiveness of a buffer depends on two factors:
  1. The relative amounts of acid and base.
  2. The absolute concentrations of acid and base.

Buffering Capacity

• Buffering capacity is the amount of acid or base that can be added to a buffer without causing a large change in pH.
• The buffering capacity increases with increasing absolute concentration of the buffer components.
• A concentrated buffer can neutralize more added acid or base than a dilute buffer.

Effectiveness of Buffers

• As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves.
• Buffers that need to work mainly with added acid generally have [base] > [acid].
• Buffers that need to work mainly with added base generally have [acid] > [base].
• A buffer will be most effective when the [base]:[acid] = 1, meaning equal concentrations of acid and base.
• A buffer will be effective when 0.1 < \frac{[base]}{[acid]} < 10.
• A buffer will be most effective when the [acid] and the [base] are large.

Buffering Range

• We have said that a buffer will be effective when 0.1 < \frac{[base]}{[acid]} < 10.
• Substituting into the Henderson–Hasselbalch equation we can calculate the maximum and minimum pH at which the buffer will be effective.
• Lowest pH: pH = pKa + log(0.1) = pKa -1
• Highest pH: pH = pKa + log(10) = pKa + 1
• Therefore, the effective pH range of a buffer is pK_a ± 1.
• When choosing an acid to make a buffer, choose one whose pK_a is closest to the pH of the buffer.

Concept Check

• Which of the following is true?
  • The correct answer is: A buffer is most resistant to pH change when [acid] = [conjugate base].